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Math Help - going in circles

  1. #1
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    Question going in circles

    Hi folks,

    I am trying to solve the following question:

    Find the equations of the two circles which pass through the point (2,0) and have both the y-axis and the line y - 1 = 0 as tangents.

    This seems like an easy question because you can do it by scale drawing. The first (small) circle can't really go anywhere else - see attachment! But assuming you had to do it by calculation.

    To find the equation of a circle we need 3 points on the circumference. We have one supplied directly (2,0) then we have the general equation:

    <br />
x^2 + y^2 + 2fx + 2gy + c = 0                                  (1)<br />

    plus the two tangents:

    x = 0

    y = 1

    substituting (2,0) gives:

    4 + 4f + c = 0

    and substituting x = 0 into the general equation gives:

    y^2 + 2gy -4(f + 1) = 0

    a quadtratic in y. Since the tangent touches the circle we expect a repeated root i.e. b^2 - 4ac = 0

    4g^2 + 4.1.4(f + 1) = 0

    g^2 + 4f + 4 = 0    (4)


    then substitute y = 1 into the general equation:

    x^2 + 1 + 2fx + 2g -4(f + 1) = 0

    x^2 + 2fx + 2g + 1 - 4(f + 1) = 0

    again setting b^2 - 4ac = 0 gives:

    <br />
4f^2 + 16f - 8g + 12 = 0                                             (5)<br />


    Equations (4) and (5) are correct. What I mean is that when you substitue the correct answers in, they give consistent results but how do you solve them? It seems clear that I am on the wrong track. I then realised that the equation y = -x + 1 must be a diameter to both circles so I tried substituting this into the general equation (1) to solve for the intersection points. It yielded a result for x with many terms in f and g that I couldn't reduce to anything like the correct answer.

    Any help here would be appreciated!
    best regards
    Attached Thumbnails Attached Thumbnails going in circles-m16f7.bmp  
    Last edited by s_ingram; July 23rd 2009 at 01:50 AM.
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  2. #2
    MHF Contributor
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    Hi

    AS you said the line y=-x+1 is a diameter of the circles
    The centers are on this line
    Let C (xC,yC) one center of the possible circles
    yC = -xC + 1

    The distance between C and the point M(2,0) is such that CM² = (xC-2)²+yC² = 2xC² - 6 xC + 5
    This distance is the same as the one between C and the y-axis, which is |xC|
    Therefore : 2xC² - 6 xC + 5 = xC²
    xC² - 6 xC + 5 = 0
    (xC - 1)(xC - 5) = 0
    xC = 1 or 5
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  3. #3
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    Quote Originally Posted by s_ingram View Post
    Hi folks,

    I am trying to solve the following question:

    Find the equations of the two circles which pass through the point 2,0) and have both the y-axis and the line y-1=0 as tangents.

    ...
    1. I used this equation of a circle:

    (x-a)^2+(y-b)^2=r^2 with the center at C(a, b) and the radius r.

    2. You already know 3 points of the circle: (2, 0), (0, 1-r), (a, 1)

    3. Substitute the coordinates of these points into the equation of the circle. You'll get a system of equations. Solve this system for a, b and r.

    \left|\begin{array}{rcl}(2-a)^2+(0-b)^2&=&r^2 \\ (0-a)^2+(a+1-b)^2&=&r^2 \\ (a-a)^2 + (1-b)^2&=&r^2 \end{array}\right.

    4. After some transformations and calculations you'll get:

    Circle #1: (1,0); r=1
    Circle #2: (5,-4); r=5
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  4. #4
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    Hello, s_ingram!

    Find the equations of the two circles which pass through the point (2,0)
    and have both the y-axis and the line y\,=\, 1 as tangents.
    Code:
            |
            |         R
          - + - P - * o * - - - y=1
        ----+---o-----:-----*----------
            | * 2     :       *
            |*        :        *
            |         :
            *         :         *
          Q o - - - - o C       *
            *       (x,y)       *
            |
            |*                 *
            | *               *
            |   *           *
            |       * * *
            |

    The center of the circle is C(x,y)

    It passes through the point P(2,0)
    It is tangent to the y-axis at Q(0,y)
    It is tangent to y = 1 at R(x,1)

    The radius is: . CP \:=\: CQ \:=\: CR \:=\: r


    We have: . x \:=\:r

    Since CR \:=\:1-y \:=\:r\:=\:x, then: .  y \:=\:1-x

    Since CP \,=\,R, we have: . \sqrt{(x-2)^2 + (y-0)^2} \:=\:r

    . . which becomes: . \sqrt{(x-2)^2 + (1-x)^2} \:=\:x

    Then: . (x-2)^2 + (1-x)^2 \:=\:x^2

    . . which simplifies to: . x^2 - 6x + 5 \:=\:0

    . . which factors: . (x-1)(x-5) \:=\:0

    . . and has roots: . x \:=\:1,5

    . . . . . . . . Then: . y \:=\:0,\text{-}4


    One circle has center (1,0) and r = 1\!:\quad \boxed{(x-1)^2 + y^2 \:=\:1}

    The other has center (5,\text{-}4) and r = 5\!:\quad \boxed{(x-5)^2 + (y+4)^2 \:=\:25}

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  5. #5
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    Excellent! Thanks a lot guys.
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